Question

Qi) Here is a BBC unit lattice. Using the full sphere model, calculate the relative amount of area that is occupied by atoms with their center in the plane shown below (diagonal across the body). This is called the planar packing factor, equivalent to the atomic packing factor in 3D. Also mark the octahedral sites present on the plane. Bec Q2) Zinc has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.85. If the radius of the Zn atom is 0.1332 nm. a) deterniine the unit cell volume (You can still assume a 2r the 2D planes are just not closely packed anymore) b) calculate the density of Zn and compare it with the literature value. Explain why the values might be different

please could you explain how to do 1 and 2. thank you

Answer 1

Ans 2a - given- radius of Zn atom () = 0.1332 nm c/a = 1.85 ar er substituting the value of er in equ. O a = 2 2X.1332 = 2664 X10 am c/a= 1.85 c=1.850 a 1.85 x 0.2664X169 The volume of the HCP unit cell ) = 38a²c substituting the values V= 383 x63864816-9 x 1-85X2644X154

V= 0.181810-27 2 = 0.0908X16-27 m3 0.0908 x 10 Ans: - Unit cell volume is

Ans 26 - Density of Zinc (e) Effective number of atoms in unit cell of zince n=6 as we know = nm Na XV where n=6, M = 65.37 Na = 6:023881026 V= 0.090881077 m3 substituting the value e= 6X 65.37 6.02388/0768.890881627 = 6X 65.37 0.5470 X61 = 392.22 0.05470 = 7170.38 Kg/m3 Density of zinczu) is 7170.38 kg / m3 Ans -

Ans:- Atomic packing factor - Natom atom i Vanit cell - Number of atoms in unit cell & Volume of at Volume of unit cell atoms in Body-centred cubic (BCC) There are only Natom = 2 We consider that atomic structure to be spherical in shape so, Vatom = volume of sphere - IT 3 de- radius of atom where 즈 ac a B

According to lythagoras thorem AC² = AB + BC? and A² = AC² + CO² From diagram 0 AD = retenerten = 42 As we know that edge of cube are equal AB= BC- CD = a Substituting the value of AC² - a²+q² AB and BC in eq. € = Ral Substituting the value of Ace in eq. AD² = 22²+CD² Subsituting the value of AD and CD in equ 4 (ur) = 22²+ a2 16 r² = 3a² a = 162² a= ur

As we know that unit cell in a cake. so volume of unit cell= Volume of cule Vunitcell - a Vunit cell = (ur) Substituting the values in eq. APF Bcc = Natom Vatem Vunit cell 4 kr = 2 x - 7X4 XT X² X & XB 2 4X44Y X A8 xz - T 3 8 0.6802 Atomic packing factor is 0.6802 Ans 1 -

Octa hedral sites present on the plane.